What is the Definition and Mathematical Explanation of Density of States?

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SUMMARY

The density of states (DOS) quantifies the number of quantum states with energy within a specific interval, defined mathematically as g(E)=∑sδ(E−Es). For free particles in a volume V, the DOS for wavevector k is expressed as g_k=V/(2π)³. This formulation arises from applying periodic boundary conditions to free waves, allowing for the integration of states over k-space. The relationship between energy E and wavevector k is crucial, particularly when considering particles with effective mass m*, leading to the equation g(E)=4πV/(2π)³ m*√(2Em*).

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  • Understanding of quantum mechanics principles
  • Familiarity with wavefunctions and periodic boundary conditions
  • Knowledge of mathematical integration in multiple dimensions
  • Basic concepts of effective mass in solid-state physics
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Definition/Summary

This term most commonly refers to the number of quantum states having energy within a given small energy interval divided by that interval.

Equations

[tex] g(E)=\sum_{s}\delta(E-E_s)[/tex]
[tex] N=\int dE g(E)[/tex]

The "density of states" need not (but it most often does) refer to states per energy interval. For example, for free particles in a box of volume [itex]\mathcal{V}[/itex], the density of states for a given wavevector [itex]\mathbf{k}[/itex] (rather than energy) is a constant:
[tex] g_{\mathbf{k}}=\frac{\mathcal{V}}{{(2\pi)}^3}.[/tex]
The above equation is the basis for the well-known replacement
[tex] \sum_{\mathbf{k}}(\ldots)\to\int \mathcal{V}\frac{d^3 k}{{(2\pi)}^3}(\ldots)[/tex]

Extended explanation

The density of states
[tex] g_{\mathbf{k}}=\frac{\mathcal{V}}{{(2\pi)}^3}\;,[/tex]
results from applying periodic boundary conditions to free waves in a box of volume [itex]\mathcal{V}[/itex] and counting. Thus
[tex] \delta N = d^3 k g_{\bf k}=d^3 k\frac{\mathcal{V}}{{(2\pi)}^3}\;.[/tex]

If the energy E only depends on the magnitude of [itex]\mathbf{k}[/itex], E=E(k), then we may also write
[tex] \delta N = d k k^2 \frac{4\pi \mathcal{V}}{{(2\pi)}^3}<br /> =<br /> \frac{4\pi\mathcal{V}}{{(2\pi)}^3}dE \frac{k^2}{v}\equiv dE g(E)\;,[/tex]
where
[tex] v=\frac{dE}{dk}\;,[/tex]
is the velocity.

For the case where momentum is carried by particles with an effective mass [itex]m^*[/itex] we have
[tex] k=m^*v\;,[/tex]
and
[tex] g(E)=\frac{4\pi \mathcal{V}}{{(2\pi)}^3}m^*\sqrt{2 E m^*}\;.[/tex]

* This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!
 
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